Problem: A particle moves along the $x$ -axis. The function $v(t)$ gives the particle's velocity at any time $t>0$ : $v(t)=t^4-2t^3-4t$ What is the particle's velocity $v(t)$ at $t=1$ ? $v(1)=$
Explanation: In the first part, we need to find the particle's velocity. We have a function for the particle's velocity, so all we have to do is evaluate the function. In the second part, we need to find the particle's acceleration. Since acceleration is the rate of change of velocity, we need to find the derivative of $v(t)$. In other words, if $a(t)$ gives the particle's acceleration at any time $t>0$, then $a(t)=v'(t)$. First, let's evaluate $v(1)$ : $\begin{aligned} v({1})&=({1})^4-2({1})^3-4({1}) \\\\ &=-5 \end{aligned}$ Notice that the velocity is negative. This means the particle is moving to the left. Now, let's differentiate $v(t)$ to find $a(t)$ : $\begin{aligned} a(t)&=v'(t) \\\\ &=\dfrac{d}{dt}[t^4-2t^3-4t] \\\\ &=4t^3-6t^2-4 \end{aligned}$ To find the particle's acceleration at $t=1$, we need to evaluate $a(1)$. $\begin{aligned} a({1})&=4({1})^3-6({1})^2-4 \\\\ &=-6 \end{aligned}$ Since the particle is accelerating toward the left while moving toward the left, we know the particle is speeding up. The particle's velocity at $t=1$ is $-5$. The particle's acceleration at $t=1$ is $-6$. At $t=1$, the particle is speeding up.